Let $a:=\alpha$ and $u:=\frac1{5\theta}$. We shall assume that $0<u<1/2$; otherwise, the problem does not seem to make much sense. We need to compute 
\begin{equation}
	\inf_{u\in(0,1/2)}\sup_{a\in[u,1/2]}F(u,a), 
\end{equation}
where
\begin{equation*}
	F(u,a):=\begin{cases}
	F_1(u,a)&\text{ if }1/5<a\le1/2,u\le a,\\
	F_2(u,a)&\text{ if }u\le a\le1/5,
	\end{cases}  
\end{equation*}
\begin{equation}
	F_1(u,a):=\frac{30 (a-1) \log (1-a)-30 a \log (a)-410 u+9}{30 (a-1)},
\end{equation}
\begin{equation}
	F_2(u,a):=\frac{H(u,a)}{60(a-1)^2 a},  
\end{equation}
\begin{multline}
	H(u,a):=-60 a^3 \log (a)-820 a^2 u-213 a^2+60 a^2 \log (a)+820 a u+78 a \\ 
	+60 (a-1) a \log
   \left(\frac{1}{2} \left(\frac{1}{a}-1\right)\right)+60 (a-1)^2 a \log (1-a)+15. 
\end{multline}
So, the infsup in question is
\begin{equation}
	\inf_{1/5<u<1/2}M_{12}(u)\wedge\inf_{0<u\le1/5}(M_{11}\vee M_2(u)), 
\end{equation}
\begin{equation}
	M_{11}:=\sup_{1/5<a\le1/2}F_1(u,a),\quad
	M_{12}(u):=\sup_{a\in[u,1/2]}F_1(u,a),\quad
	M_2(u):=\sup_{a\in[u,1/5]}F_2(u,a). 
\end{equation}
 
For $F_1(a):=F_1(u,a)$, let $DF_1(a):=F_1'(a)(1-a)^2$. Then $DF_1(a)=-3/10 + 41 u/3 + \log a$ is increasing in $a$. So, $DF_1(a)$ (and hence $F_1'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{align}
	M_{11}&=\sup_{1/5<a\le1/2}F_1(u,a)=F_1(u,1/5)\vee F_1(u,1/2),\\ 
	M_{12}(u)&=\sup_{u\le a\le1/2}F_1(u,a)=F_1(u,u)\vee F_1(u,1/2). 
\end{align}

For $F_2(a):=F_2(u,a)$, let $DF_2(a):=F_2'(a)(1-a)^2$. Then $DF_2'(a) 2 (1-a)^2 a^3=1 - 5 a + 11 a^2 - 17 a^3 + 2 a^4>0$ for $a\in[0,1/5]$. So, $DF_2(a)$ is increasing in $a\in[0,1/5]$, and so, $DF_2(a)$ (and hence $F_2'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{equation}
	M_2(u)=\sup_{a\in[u,1/5]}F_2(u,a)=
	F_2(u,u)\vee F_2(u,1/5)\quad\text{if }0<u\le1/5. 
\end{equation}



So, the infsup in question is
\begin{equation}
	\inf_{1/5<u<1/2}[F_1(u,u)\vee F_1(u,1/2)]
	\bigwedge\inf_{0<u\le1/5}[F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)], 
\end{equation}
We have 

>**Lemma 1.** $F_1(u,u)\le F_1(u,1/2)$ if $1/5<u<1/2$. 

>**Lemma 2.**  
\begin{equation}
	\inf_{1/5<u<1/2}F_1(u,1/2)=F_1(1/5,1/2)=-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots.
\end{equation}

>**Lemma 3.** 
\begin{equation}
	\inf_{0<u\le1/5}F_2(u,u)=\inf_{0<u\le1/5}g(u)=3.7373\ldots>3.4803\dots.
\end{equation}

These lemmas will be proved at the end of this answer. Since 
$F_1(u,1/5)\vee F_1(u,1/2)\vee F_2(u,u)\vee F_2(u,1/5)\ge F_2(u,u)$, 
it follows from these lemmas that 

>the infsup in question is $-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots$, "attained in the limit" at $u=1/5$ (cooresponding ot $\theta=1$) and $a=1/2$. 

-------------

It remain to prove Lemmas 1,2,3. 

*Proof of Lemma 1.* Indeed assume that 
$1/5<u<1/2$. We have to check that
\begin{equation}
	d_1(u):=F_1(u,1/2)-F_1(u,u)\ge0. 
\end{equation}
Let $Dd_1(u):=d1'(u) 30 (1-u)^2=419 - 1640 u + 820 u^2 - 30 \log u$. Then $(Dd_1)'(u)=10(-3 - 164 u + 164 u^2)/u<0$. So, $Dd_1(u)$ is decreasing. So, $Dd_1(u)$ (and hence $d_1'(u)$) can change the sign only from $+$ to $-$. Also, $d_1(1/5)>0=d_1(1/2)$. So, $d_1(u)\ge0.\quad$ $\Box$

*Proof of Lemma 2.* 
We have $F_1(u,1/2)=-3/5 + 82 u/3 - \log 4$, which is increasing in $u$. So,  
\begin{equation}
	\inf_{1/5<u<1/2}F_1(u,1/2)=-3/5 + 82 (1/5)/3 - \log 4=3.4803\dots.\quad \Box
\end{equation}


*Proof of Lemma 3.* 
For $0<u\le1/5$, consider  $g(u):=F_2(u,u)$ and then $Dg(u):=g'(u)(1 - u)^2$. Then $(Dg)'(u)$ is a simple rational function of $u$, which is $\ge0$. So, $Dg(u)$ is increasing in $u$. So, $Dg(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find 
\begin{equation}
	\inf_{0<u\le1/5}F_2(u,u)=\inf_{0<u\le1/5}g(u)=3.7373\ldots>3.4803\dots. \quad \Box
\end{equation}